# Rolle's Theorem

1. Mar 30, 2009

### msc8127

1. The problem statement, all variables and given/known data
determine whether rolle's theorem can be applied to f on the closed interval [a,b]. If Rolle's theorem can be applied, find all values of c in the open interval (a,b) such that f'(c) = 0

2. Relevant equations
f(x) = (x-1)(x-2)(x-3) with the interval being [1,3]

3. The attempt at a solution
the function is continuous on [1,3] and differentiable on (1,3)

f(1) = 0 and f(3) = 0 so f(1) = f(3)

f'(x) = 3x^2 - 12x + 11

now I need to find the point in [1,3] with slope 0 so I set f'(x) = 0.

I know from here i'm supposed to get c values of (6 - sqrt 3) / 3 and (6 + sqrt 3) / 3 however the algebra to get me to this point is eluding me.

Can someone please point of the obvious for me?

Thanks

2. Mar 30, 2009

### CompuChip

It is simply a quadratic equation:
$$3x^2 - 12x + 11 = 0$$
You can solve it using your standard toolbox, e.g. by applying the quadrature formula:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$.

3. Mar 30, 2009

### brendan

use use the Quadratic equation to solve for x

x = (-b +/- sqrt(b^2 - 4ac)) / 2a

your equation is f'(x) 3x^2 - 12x + 11

Set for zero

3x^2 - 12x + 11=0

So.

a = 3 b = -12 c= 11

x = (12+/- Sqrt(144 - (4*3*11)))/6

x = (12 +/- Sqrt(12))/6

x = (Sqrt(3)+6)/3 or -(Sqrt(3)-6)/3